  # Country Boy

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1. Your title is confusing. Saying "life didn't start on Earth" is not the same as saying "life didn't only start on earth"!
2. When I first saw "Earth's satellite and its first space speed" I immediately assume that the "Earth's satellite" referred to the moon! Then I opened it and saw "the artificial satellite" my first thought was "the"? There are many satellites- "an artificial satellite". In any case, you can calculate the speed at which any satellite, at a given height,, must travel to maintain that height. And a "geosynchronous" satellite must travel at a speed that completes a full orbit (circle) in 24 hours.
3. You have two different ranking methods with rank A and B in different orders. To be able to determine an order for A and B you will have to decide on how to weight the two methods. That will depend on which you think is more "accurate". For example if you think method 1 is twice as accurate as method 2, you would give 1 a weight of 2/3 and 2 a weight of 1/3 (because 2/3 is twice 1/3 and they add to 1). So you would rank A as (2/3)(.7)+ (1/3)(.5)= 19/3 and would rank B as (1/3)(.5)+ (1/3)(.7)= 12/3. A is ranked higher than B. But if you thought 2 was 50% more accurate than 1 you would give 1 a weight .4 and 2 a weight of .6 (.6= (1.5)(.4) and .4+ .6= 1). So now you would rank A as .4(.7)+ .6(.5)= .28+ .30= .58 and B as .4(.5)+ .6(.7)= .20+ .42= .62. Now B is ranked higher than A. Of course, if you believe they are equally accurate, you would weight both 1 and 2 as 0.5. Now A is given the combined ranking of 0.5(0.7)+ 0.5(0.5)= .35+ .25= 0.6 and B is given the combined rating of 0.5(0.5)+ 0.5(0.7)= .25+ .35= 0.6 and both are equal in rating. (Note that if methods 1 and 2 gave the same ranking for A and B, no matter what the actual values are, the combined ratings, using any weights, would give the same ranking.)
4. Is this a "Calculus" problem? The "standard" way to calculate (or even define) "speed at an instant" is the derivative with respect to time. The shark is a constant 90 feet from the shore and moving along the shore at a constant 30 feet per second so its distance down the shore from the lifeguard station is 30t feet where t is the time, in seconds, since the shark passed the lifeguard station. By the Pythagorean theorem, the straight line distance from the lifeguard station to the shark, at time "t", is given by $D^2= 90^2+ (30t)^2[math]. Differentiating both sides, [math]2D (dD/dt)= 2(30)(30t)= 1800t$. The problem asked for the distance when the shark is D= 150 feet past the lifeguard station but we need to also calculate t at that point. Use $D^2= 150^2= 22500= 90^2+ (30t)^2= 8100+ 900t^2$. So we need to solve $900t^2= 22500- 8100= 14400$ or $t^2= 14400/900= 16$. t= 4 and D= 150 so $2D(dD/dt)= 1800t$ becomes $300 (dD/dt)= 7200$. The rate at which the straight line distance from the lifeguard station to the shark is increasing, when the shark is 150 feet from the lifeguard station, is $dD/dt= 7200/300= 24$ feet per second.
5. If I understand what you have written, $3975,3- F_{35}cos(26,6)+ F_{32}cos(26,6)= 0$ so, adding $F_{35}cos(26,6)- 3975,3$ to both sides, $F_{32}cos(26,6)= F_{35}cos(26,6)- 3975,3$ and then, dividing both sides by cos(26,6), $F_{32}= F_{35}- \frac{3975,3}{cos(26,6)}$.You are told that $F_{35}= 50326,9$ and can use a calculator to determine that cos(26,6)= 0.894. Put those numbers in and do the arithmetic!
6. I would write every vector in terms of components. T= <-22900, 0>, Weng= <0, -24525>. The sum of those two forces is <-22900, -24525> and the two forces, "45" and "35" must offset that. Taking the force in vector "45" to be "x" and in vector "35" to be "y" we have "45"= <0, x> and "35"= <0.6y/1.34, 1.2y/1.34>= <0.45y, 0.90y>. The sum of those is <0.45y, x+ 0.90y> and that must equal <22900, 24525> so we have 0.45y= 22900 and x+ 0.90y= 24525. From the first y= 22900/0.45= 50,889. Then the second is x+ 45800= 24525 so x= 24525- 45800= -21275. So vector "45" is <0, -21275> and "35" is <50899(.45), 50899(.90)>= <22904.55, 45809.1>. The others are done the same way.
7. Any linear function of x can be written f(x)= ax+ b for numbers a and b. You are told that f(2002)= 120= 2002a+ b and f(2013)= 212= 2013a+ b. Solve the equations 2002a+ b= 120 and 2013a+ b= 212 for a and b.
8. No, because such a "set of points" would not be connected so not a line in the geometric sense.
9. When I click on that link I get a message that says "you do not have permission to do that"!
10. Well, you can't define or calculate the derivative of a function using only the four arithmetic operations. I would have thought you would have learned that in whatever course you were introduced to the derivative. You have to have the concept of a "limit" as well: $$\frac{df}{dx}(a)= \lim_{h\to 0} \frac{f(a+ h)- f(a)}{h}$$.
11. If I am understanding this correctly, a fraction represents a terminating decimal (so there is an "end") if and only if the denominator has factors of "2" and "5" only. How many decimal places is a little harder. 1/2= .5, 1/4= 1/2^2= .25, 1/8= 1/2^3= .125, 1/10= 1/(2*5)= 0.1, 1/16= 1/2^4= .0625, 1/20= 1/(2^2*5)= .05, etc. Do you see a pattern?
12. Okay, your teacher clearly expects you to know what the "LCM" method is! Do you? (My first thought was "Least Common Multiple" but that might not apply here.)
13. I don't understand the purpose of this thread. Have you ever actually taken a class in algebra?
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